Low current continuity checker

Question

Let me preface this by saying I have minimal electronics experience and reading data sheets is slightly confusing, but I get a gist of how most components work.

I am making a lower current continuity checker. What I mean by this is the circuit I'm checking will must have uA levels of current flowing through it. I have made a simple circuit with two PN2222a NPN bi-polar transistors and am using the base to read continuity and activating a small load of 180mA.

The current measurements I am getting are as follows.

The issue is that the transistors do not have a solid "break point" where they are on or off with such a low input voltage/current. I have considered using a comparator but don't want to run into the same issue where the comparator is not sensitive enough to decern the difference of 2 uA.

The goal is to keep this circuit as small as possible. I would be happy with a continuity checking range of 0 Ω to 5 kΩ.

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Answers:

The answers from @Spehro Pefhany and @ Simon Fitch contributors are good and solve the issue I was having. I do thoroughly enjoy @spehro's solution because you made a comparator with BJTs. It made me think more on the function of BJT's and a more dynamic usage of them to include helping me understand comparators as well.

Both circuits can be adjusted to different OHM ranges, have uAs of current through the test lead, will work on 3v, and involve minimal components.

Thanks again for all the help! If I could say both solved the problem I would.

Answer 1

I'll assume your battery is 3V.

You don't say how many microamps, but judging by your own design I'm guessing \$I=\frac{3V}{150\rm{\ k\Omega}} = 20\rm{\ \mu A}\$.

The voltage developed across a 5kΩ resistance would be \$V=IR=20\rm{\ \mu A}\times 5\rm{\ k\Omega} = 100\rm{\ mV}\$. You need your output to flip only in the last few millivolts of that range, as the resistance crosses the 5kΩ threshold, which will require a voltage gain of say \$\frac{3\rm{\ V}}{10\rm{\ mV}}=300\$.

This will be difficult to achieve using a transistor in common-emitter configuration, because the 0.7V (or 1.4V for a darlington pair) threshold is uncertain, and even the most precise biasing won't help. You need some emitter degeneration to provide predictability over time, temperature and component tolerances.

On top of that, your design constraint of having 180mA output capability means that the required current gain would be well in excess of \$\frac{180\rm{ mA}}{20\rm{ \mu A}} = 9000\$.

This circuit overcomes those sources of uncertainty by biasing itself, and will produce extremely high gain:

^{simulate this circuit – Schematic created using CircuitLab}

Ignore D1, it's just there to constrain I2's voltage. High gain is achieved due to current sources I1 and Q2 competing to control \$V_{OUT}\$. Even the tiniest imbalance in current between those two elements will send \$V_{OUT}\$ quickly to one extreme or the other.

The reason I suggest this design is that the symmetry of the two current paths means that an imbalance caused by \$R_{DUT} \ne R_{REF}\$ is what causes the output to transition. In other words, \$R_{REF}\$ determines the "threshold" resistance which will cause the output to flip.

R1 is chosen to have \$I_1\approx I_2\$. This balances the two halves, and any difference between the resistance of device under test (DUT) \$R_{DUT}\$ and \$R_{REF}\$ will upset that balance, and cause \$V_{OUT}\$ to shoot off to one extreme or the other.

This is \$V_{OUT}\$ as I sweep \$R_{DUT}\$ between zero and 10kΩ:

Current through the DUT is always small:

You can implement a current source with one more transistor, as follows:

^{simulate this circuit}

It's not a very good current source, so you should adjust R2 until you have 20μA through it, which you can verify by measuring the voltage across it, and applying Ohm's law.

A better solution would be to use a current mirror for both \$I_1\$ and \$I_2\$, which would bias and balance itself very well:

^{simulate this circuit}

Here R6 is chosen to set current down both paths, which can be calculated as follows:

$$ I_1 \approx I_2 \approx \frac{V_{SUPPLY}-1.4V}{R_6} \approx \rm{\ 16\mu} A$$

Sorry it's not as simple as you'd no doubt like, but this is the price to pay for such high gain and small currents. Having said that, this is the first design that I thought of, and there are very likely to be other simpler designs that provide similar performance.

The next step is to buffer \$V_{OUT}\$, so that your load doesn't sink or source more than a couple of microamps from OUT. You could use a darlington pair, sziklai pair, MOSFET or comparator in the usual way, and is easy now that \$V_{OUT}\$ swings almost all the way to the supply rails.

Answer 2

A simple method using an LMV761 comparator is shown here:

The comparator is chosen because it has very low input bias current (pA) and very low offset voltage (200uV typical). The latter is required in order to reliably detect differences of mV.

R4 (and the balancing R8, which can be omitted) are there to limit the voltage across the test resistance R7 to about 100mV. If you don't care about that, you can leave them out. R6 is the nominal threshold where the comparator switches. M1 switches the load.

Depending on your application you might want to add some hysteresis, some input protection, some filtering, and definitely some power supply bypass capacitance.

The maximum current through R7 is about 1uA and the maximum (open circuit) voltage is 100mV.

You could also use a "zero drift" CMOS op-amp as a comparator to get even lower input offset voltage (uV). The comparator has a relatively low gain.